$\begingroup$+1 for reminding me that I haven't played Mastermind in a while...such a fun game.$\endgroup$
– Brandon_JJan 29 '19 at 19:06

1

$\begingroup$While Mastermind technically has 4 digits, I figured it was close enough that tagging it wouldn't be too off-putting. Glad I reminded you of the game. :)$\endgroup$
– Ian MacDonaldJan 29 '19 at 21:10

9 Answers
9

Because 1 is in the same place as it was before, but the "one correct" is in the wrong place, it cannot be a right number. This means that 5 (from earlier) is definitely the last number, and also one of either 6 or 7 is correct, but in the wrong place.

[4, 3, 0]:

We know from above that 4 and 3 are both wrong, and we deduced from the last step that 1 was not a number, meaning the first line shows us that 5 and 0 are both numbers, so this actually gives us no new information on which numbers, but does tell us that 0 is still in the wrong place.

This means that we have "0 x 5" with x being either 6 or 7. From the second-last line, we know that the correct number is in the wrong place, so it can't be 6 (which was in the middle).

$\begingroup$I think it's awesome that you got to the conclusion relatively simply, but didn't get there by the much more important piece of info those first two rules gave: switching from 0 to 3 gave one less correct number - which means 0 has to be one of the three numbers (and, conversely, 3 cannot.)$\endgroup$
– KevinJan 29 '19 at 22:20

I've found the answers too complicate, so I share mine. Only three rules are needed:

501 — Two correct numbers in wrong places

So, only one of those numbers should be removed. Which one?

135 — One correct number in the right place

Since this shares 2 numbers with the first one, the 3 cannot be the
correct one. (If the 3 comes into play, the rule would give info about
2 numbers instead of one). So, from the previous rule, we should
remove only the 1 or the 5. Now we know the number has a 0 in one side,
and starts with 1 or end with 5. Only 2 chances: 1_0 or 0_5.

167 — One correct number in wrong place

This number start with 1 as the previous one, and the 1 can't be in
right and wrong place at the same time. So, we have only one
possibility now: 0_5. We are only missing the middle number and can't be the
6 because it's already in the middle and not in the right place. So
the only option left is the 7. -> 075.

0 Is right because there are two numbers right in first condition. And other one is wrong because of second condition in which 5 or 1 Has to be wrong. and From last condition 0 is selected and other 2 values got discard because of the 3rd condition. We got the position of the 0 from these two upper conditions. as 0 is wrong at second place in first condition and 0 is also wrong at last place in last condition.5 Is selected from first condition if we select 1 then we get only 2 numbers as we need 3 to make the key. 5 is also right in condition second in which we also got the right position of the 5.7 is right because of 4th condition as the other two number which is 1 is wrong due to first and second condition because we selected 5 and 6 is also wrong due to the position of 6. we need number that fits in position between 5 and 0. because 7 is at wrong place that fits at second position.As it satisfy all conditions501 — Two correct numbers in wrong places(5,0, As 1 or 5 get discard in second condition, so i took 5)135 — One correct number in the right place(5, As we can select only one that should be from first condition.)483 — All numbers are wrong(If presented above shall be discard.)167 — One correct number in wrong place(7, Both the last and first place is booked thats why only chance 7)430 — One correct number in the wrong place(0, as we took in the first step.)

a. Via rules 2 & 3 the answer contains a 1 or a 5 but not both.
b. Via (a), rule 1, and rule 2 the 1 or 5 in the answer is not in the middle.
c. Rules 2 and 4 contradict if 1 is in the number, so 1 is not in the number, therefore 5 is at the end.
d. Via (b), (c), and rule 1 the answer contains a 0 which is not in the middle, therefore 0 is at the front.
e. Via (c), (d), and rule 4, 7 is in the middle.

Rule 1) 501 — Two correct numbers in wrong places
Rule 2) 135 — One correct number in the right place
Rule 3) 483 — All numbers are wrong
Rule 4) 167 — One correct number in wrong place
Rule 5) 430 — One correct number in the wrong place

Starting from the most elimination Rule 3

Because of Rule 3, Rule 5 proves that 0 is at position 1 or 2.
But 0 can't be at position 2 because of Rule 1.

So 0 is at position 1.

Finding the 2nd digit

Because of Rule 1, and we know about 0, it has to contain a 1 or a 5, but not both.
And because of Rule 2, it can't be 1 because that position is already taken by 0.

So 5 is at position 3.

The remaining digit

Then there's only the 2nd position to figure out.
And because of Rule 4, it can't be 6.
The 1 had already been discarded when we discovered 5 is at position 3.